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Polynomials and Quadratics Higher hsn .uk.net Mathematics UNIT 2 OUTCOME 1 Polynomials and Quadratics Contents Polynomials and Quadratics 64 1 Quadratics 64 2 The Discriminant 66 3 Completing the Square 67 4 Sketching Parabolas 70 5 Determining the Equation of a Parabola 72 6 Solving Quadratic Inequalities 74 7 Intersections of Lines and Parabolas 76 8 Polynomials 77 9 Synthetic Division 78 10 Finding Unknown Coefficients 82 11 Finding Intersections of Curves 84 12 Determining the Equation of a Curve 86 13 Approximating Roots 88 HSN22100 This document was produced specially for the HSN.uk.net website, and we require that any copies or derivative works attribute the work to Higher Still Notes. For more details about the copyright on these notes, please see http://creativecommons.org/licenses/by-nc-sa/2.5/scotland/ Higher Mathematics Unit 2 – Polynomials and Quadratics OUTCOME 1 Polynomials and Quadratics 1 Quadratics A quadratic has the form ax2 + bx + c where a, b, and c are any real numbers, provided a ≠ 0 . You should already be familiar with the following. The graph of a quadratic is called a parabola . There are two possible shapes: concave up (if a > 0 ) concave down (if a < 0 ) This has a minimum This has a maximum turning point turning point To find the roots (i.e. solutions) of the quadratic equation ax2 + bx + c = 0, we can use: factorisation; completing the square (see Section 3); −b ± b2 − 4 ac the quadratic formula: x = (this is not given in the exam). 2a EXAMPLES 1. Find the roots of x2 −2 x − 3 = 0 . x2 −2 x − 3 = 0 ()()x+1 x − 3 = 0 x+=1 0 or x −= 3 0 x= −1 x = 3. 2. Solve x2 +8 x + 16 = 0 . x2 +8 x + 16 = 0 ()()x+4 x + 4 = 0 x+=40or x += 40 x=−4 x =− 4. hsn .uk.net Page 64 HSN22000 Higher Mathematics Unit 2 – Polynomials and Quadratics 3. Find the roots of x2 +4 x − 1 = 0 . We cannot factorise x2 +4 x − 1 , but we can use the quadratic formula: −±4 42 −××− 41() 1 x = 2× 1 −4 ± 16 + 4 = 2 −4 ± 20 = 2 = −4 ± 4 5 2 2 = −2 ± 5. Note • If there are two distinct solutions, the curve intersects the x-axis twice. y y x x • If there is one repeated solution, the turning point lies on the x-axis. y y x x • If b2 −4 ac < 0 when using the quadratic formula, there are no points where the curve intersects the x-axis. y y x x hsn .uk.net Page 65 HSN22000 Higher Mathematics Unit 2 – Polynomials and Quadratics 2 The Discriminant Given ax2 + bx + c , we call b2 − 4 ac the discriminant . This is the part of the quadratic formula which determines the number of real roots of the equation ax2 + bx + c = 0. y • If b2 −4 ac > 0 , the roots are real and two roots unequal (distinct). x y • If b2 −4 ac = 0 , the roots are real and equal one root (i.e. a repeated root). x y • If b2 −4 ac < 0 , the roots are not real; the no real parabola does not cross the roots x-axis. x EXAMPLE 1. Find the nature of the roots of 9x2 + 24 x + 16 = 0. a = 9 b2−4 ac = 24 2 −×× 4916 b = 24 =576 − 576 c =16 = 0 Since b2 −4 ac = 0 , the roots are real and equal. 2. Find the values of q such that 6x2 + 12 x + q = 0 has real roots. Since 6x2 + 12 x + q = 0 has real roots, b2 −4 ac ≥ 0 : a = 6 b2 −4 ac ≥ 0 b =12 122 − 4 × 6 ×q ≥ 0 c= q 144− 24q ≥ 0 144≥ 24 q 24q ≤ 144 q ≤ 6. hsn .uk.net Page 66 HSN22000 Higher Mathematics Unit 2 – Polynomials and Quadratics 3. Find the range of values of k for which the equation kx2 +2 x − 7 = 0 has no real roots. For no real roots, we need b2 −4 ac < 0 : a= k b2 −4 ac < 0 b = 2 242 − ×k ×−() 70 < c = − 7 4+ 28k < 0 28k < − 4 4 k < − 28 1 k < − 7 . 4. Show that (24k+) x2 +( 32 k +) xk +−=( 20) has real roots for all real values of k. a=2 k + 4 b2 − 4 ac b=3 k + 2 =+−()()()32k2 42 k + 4 k − 2 c= k − 2 =9kk2 + 12 +− 4()() 2 k + 44 k − 8 =9k2 + 12 k +− 4 8 k 2 + 32 =k2 +12 k + 36 =()k + 62 . Since b2 −4 ac =+( k 6)2 ≥ 0 , the roots are always real. 3 Completing the Square 2 The process of writing y= ax2 + bx + c in the form yax=( + p) + q is called completing the square . Once in “completed square” form we can determine the turning point of any parabola, including those with no real roots. The axis of symmetry is x= − p and the turning point is (− p, q ). 2 The process relies on the fact that (xp+) =+ x22 pxp + 2 . For example, we can write the expression x2 + 4 x using the bracket (x + 2)2 since when multiplied out this gives the terms we want – with an extra constant term. hsn .uk.net Page 67 HSN22000 Higher Mathematics Unit 2 – Polynomials and Quadratics 2 2 k This means we can rewrite the expression x+ kx using (x + 2 ) since this gives us the correct x 2 and x terms, with an extra constant. We will use this to help complete the square for y=3 x2 + 12 x − 3 . Step 1 Make sure the equation is in the form y=3 x2 + 12 x − 3. y= ax2 + bx + c . Step 2 Take out the x 2 -coefficient as a factor of y=3( x2 + 4 x ) − 3. the x 2 and x terms. Step 3 Replace the x2 + kx expression and y=3(() x + 22 −− 43) compensate for the extra constant. 2 =3()x + 2 −− 12 3. Step 4 Collect together the constant terms. y=3( x + 2)2 − 15. Now that we have completed the square, we can see that the parabola with equation y=3 x2 + 12 x − 3 has turning point (−2, − 15 ) . EXAMPLES 2 1. Write y= x2 +6 x − 5 in the form y=( xp +) + q . 2 y= x +6 x − 5 Note =() +2 −− You can always check x 3 9 5 your answer by =()x +32 − 14. expanding the brackets. 2 2. Write x2 +3 x − 4 in the form (x+ p) + q . x2 +3 x − 4 2 = +3 −− 9 ()x 2 4 4 2 = +3 − 25 ()x 2 4 . hsn .uk.net Page 68 HSN22000 Higher Mathematics Unit 2 – Polynomials and Quadratics 3. Write y= x2 +8 x − 3 in the form y=( xa +)2 + b and then state: (i) the axis of symmetry, and (ii) the minimum turning point of the parabola with this equation. y= x2 +8 x − 3 =()x +42 − 16 − 3 =()x +42 − 19. (i) The axis of symmetry is x = − 4 . (ii) The minimum turning point is (−4, − 19 ) . 4. A parabola has equation y=4 x2 − 12 x + 7 . (a) Express the equation in the form y=( xa +)2 + b . (b) State the turning point of the parabola and its nature. (a) y=4 x2 − 12 x + 7 =4()x2 − 3 x + 7 2 = −3 −+ 9 4(()x 2 4 ) 7 2 = −3 −+ 4()x 2 9 7 2 = −3 − Remember 4()x 2 2. If the coefficient of x2 is 3 positive then the − (b) The turning point is ( 2 , 2 ) and is a minimum. parabola is concave up. hsn .uk.net Page 69 HSN22000 Higher Mathematics Unit 2 – Polynomials and Quadratics 4 Sketching Parabolas The method used to sketch the curve with equation y= ax2 + bx + c depends on how many times the curve intersects the x-axis. We have met curve sketching before. However, when sketching parabolas, we do not need to use calculus. We know there is only one turning point, and we have methods for finding it. Parabolas with one or two roots • Find the x-axis intercepts by factorising or using the quadratic formula. • Find the y-axis intercept (i.e. where x = 0). • The turning point is on the axis of symmetry: y y x O O x The axis of symmetry is halfway A repeated root lies on the axis between two distinct roots. of symmetry. Parabolas with no real roots • There are no x-axis intercepts. • Find the y-axis intercept (i.e. where x = 0). • Find the turning point by completing the square. EXAMPLES 1. Sketch the graph of y= x2 −8 x + 7 . Since b2 −4 ac =−( 8)2 −××> 4170 , the parabola crosses the x-axis twice. The y-axis intercept (x = 0) : The x-axis intercepts ( y = 0): y =(0)2 − 80( ) + 7 x2 −8 x + 7 = 0 = 7 ()()x−1 x − 7 = 0 − = or x −7 = 0 ( ) x 1 0 0, 7 . x =1 x = 7 (1, 0 ) (7,0 ) . The axis of symmetry lies halfway between x =1 and x = 7 , i.e. x = 4 , so the x-coordinate of the turning point is 4. hsn .uk.net Page 70 HSN22000 Higher Mathematics Unit 2 – Polynomials and Quadratics y We can now find the y-coordinate: y= x2 −8 x + 7 y =(4)2 − 84( ) + 7 =16 − 32 + 7 7 = − 9. x O 1 7 So the turning point is (4, − 9 ). ( − ) 4, 9 2. Sketch the parabola with equation y=− x2 −6 x − 9 . Since b2 −4 ac =−( 641)2 −×−×−()() 90 = , there is a repeated root.
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